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(D) Given: $0.5\%$ solution of $KCl$ means $0.5 \, g$ of $KCl$ in $99.5 \, g$ of water.Molality $(m) = \frac{	ext{mass of solute}}{	ext{molar mass o

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$99$

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(D) Given: $0.5\%$ solution of $KCl$ means $0.5 \, g$ of $KCl$ in $99.5 \, g$ of water.Molality $(m) = \frac{	ext{mass of solute}}{	ext{molar mass of solute}} 	imes \frac{1000}{	ext{mass of solvent in } g} = \frac{0.5}{74.6} 	imes \frac{1000}{99.5} \approx 0.0673 \, mol \, kg^{-1}$.Using the formula for depression in freezing point: $\Delta T_f = i 	imes K_f 	imes m$.$0.24 = i 	imes 1.80 	imes 0.0673$.$i = \frac{0.24}{1.80 	imes 0.0673} \approx 1.981$.For $KCl ightarrow K^+ + Cl^-$,the number of ions $(n) = 2$.Degree of dissociation $(\alpha) = \frac{i - 1}{n - 1} = \frac{1.981 - 1}{2 - 1} = 0.981$.Percentage dissociation $= 0.981 	imes 100 = 98.1\%$.The nearest integer is $98\%$ (or $99\%$ depending on rounding precision,here $98$ is closer).

$A$ $0.5\%$ solution of potassium chloride was found to freeze at $-0.24^{\circ} C$. The percentage dissociation of potassium chloride is .... (Nearest integer)
(Molal depression constant for water is $1.80\, K\, kg\, mol^{-1}$ and molar mass of $KCl$ is $74.6\, g\, mol^{-1}$)

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$A$ $0.5\%$ solution of potassium chloride was found to freeze at $-0.24^{\circ} C$. The percentage dissociation of potassium chloride is .... (Nearest integer)(Molal depression constant for water is $1.80\, K\, kg\, mol^{-1}$ and molar mass of $KCl$ is $74.6\, g\, mol^{-1}$)